In section 5.6.1 of Evans' PDE text, Theorem 3 states:
Assume $U$ is a bounded open subset of $\mathbb{R}^n$. Suppose $u\in W_0^{1,p}(U)$ for some $1\leq p<n$. Then we have the estimate $$||u||_q\leq C||Du||_p$$ for each $q\in [1,p^*]$ where $p^*>p$ is the Sobolev conjugate $$p^*=\frac{np}{n-p}$$ and the constant $C$ depends only on $p,q,n$ and $U$.
So far so good. But then, Evans casually further states:
In particular, for all $1\leq p\leq \infty$ $$||u||_p\leq C||Du||_p.$$
Now, I have a couple concerns. First, is the above inequality even true for $p=\infty$? I confess I haven't put in too much thought into it, but I ask because wikipedia (https://en.wikipedia.org/wiki/Poincar%C3%A9_inequality) and a few other sources I've found seem to specifically exclude $p=\infty$.
The more pressing question I have is the following: Does the "special case" (the statement following "In particular") follow directly from the main statement in some trivial way? Our function $u$ has to be in $W_0^{1,p}(U)$ for some $p$ such that $1\leq p<n$, where $n$ is fixed. So, using the fact that $p<p^*$, the first statement would certainly prove the special case $||u||_p\leq C||Du||_p$ for $p<n$. But what about $\infty>p\geq n$? Does this case also follow immediately? I don't doubt that it's true, but I'm confused because Evans makes it seem like it's a trivial corollary.
I think the case $n\leq p<\infty$ can be shown in the following way: Fix $n\leq p<\infty$, and suppose $u\in W_0^{1,p}$. Next choose $1\leq q<n$ so that $q^*>p$, this selection being possible since $q^*=\frac{nq}{n-q}\to\infty$ as $q\to n^-$. Then, for this $q$ we certainly have $$||u||_{q^*}\leq C||Du||_q$$ Then, since $p<q^*$ and $q<p$, we get the desired inequality $$||u||_p\leq C'||Du||_p$$ because of the embedding relation $L^b(U)\subset L^a(U)$ whenever $a\leq b$ and $U$ bounded.
First, is this argument even correct? Second, is this the "simplest" argument? (if it is, then it doesn't seem exactly trivial enough to wave it off as an immediate consequence of the main statement above without proof..)
Yes it is true also for $p=\infty$. If you extend $f$ to be zero outside $U$ you have a Lipschitz function so you can use the fundamental theorem of calculus on segments parallel to the axes, say $$f(x)=f(y_1,x_2,\ldots,x_n)+\int_{y_1}^{x_1}\partial_1 f(t,x_2,\ldots,x_n)\,dt=0+\int_{y_1}^{x_1}\partial_1 f(t,x_2,\ldots,x_n)\,dt,$$ where $y_1$ is so large that $f(y_1,x_2,\ldots,x_n)=0$. So you get $\Vert f\Vert_\infty\le \text{diam}\,U \Vert \nabla f\Vert_\infty$. Your proof works. Otherwise, I would just prove the inequality directly. From what I wrote above, by Holder's inequality you get $$|f(x)|\le \int_{y_1}^{x_1}|\partial_1 f(t,x_2,\ldots,x_n)|\,dt \le |x_1-y_1|^{1/p'}\left(\int_{y_1}^{x_1}|\partial_1 f(t,x_2,\ldots,x_n)|^p\,dt\right)^{1/p}.$$ Raise to the power $p$ and integrate over $U$ to get $$\int_U |f(x)|^pdx\le (\text{diam}\,U)^{p/p'} \int_U\int_{y_1}^{x_1}|\partial_1 f(t,x_2,\ldots,x_n)|^p\,dtdx.$$ Use Fubini and the fact that $\nabla f=0$ outside $U$.